Question

Integrate \((x^2+9) e^{2x} dx\)

(a) \(\frac{e^{x}}{2} (x^2+x-\frac{39}{4})+C\)

(b) \(\frac{e^{2x}}{2} (x^2+x-\frac{35}{4})+C\)

(c) \(\frac{e^{2x}}{2} (x^2+x-\frac{48}{4})+C\)

(d) \(\frac{e^{x}}{2} (x^2+x-\frac{25}{4})+C\)

I have been asked this question during an interview.

I'd like to ask this question from Integration by Parts in division Integrals of Mathematics – Class 12

Answer 1

Correct answer is (b) \(\frac{e^{2x}}{2} (x^2+x-\frac{35}{4})+C\)

For explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get

\(\int (x^2+9) \,e^x \,dx=(x^2+9) \int e^{2x} \,dx-\int (x^2+9)’\int e^{2x} \,dx\)

=\(\frac{(x^2+9) \,e^{2x}}{2}-\int 2x \frac{e^{2x}}{2} \,dx\)

Again, applying integration by parts for integrating \(\int \,x \,e^{2x} \,dx\)

\(\int \,x \,e^{2x} \,dx=x\int e^{2x} \,dx-\int (x)’ \int \,e^{2x} \,dx\)

=\(\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx\)

=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\).

∴\(\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)

=\(\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})\)

=\(\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C\).